The sum of three consecutive integers is equal to their product. How many such possibilities are there?

The correct option is Only three.

Let us denote the three consecutive integers as $n-1$, $n$, and $n+1$, where $n$ is an integer.

According to the problem statement, the sum of these three consecutive integers is equal to their product. We can set up this relationship mathematically:
$(n-1)+n+(n+1)=(n-1)·n·(n+1)$

First, simplify the left side of the equation by adding the terms together:
$(n-1)+n+(n+1)=3n$

Next, simplify the right side of the equation using the difference of squares identity, $(n-1)(n+1)=n^2-1$:
$(n-1)·n·(n+1)=n(n^2-1)=n^3-n$

Now, equate the simplified left side and right side:
$3n=n^3-n$

Rearrange the terms to form a polynomial equation set to zero:
$n^3-4n=0$

Factor out the common term $n$ from the expression:
$n(n^2-4)=0$

Further factor the quadratic part using the difference of squares:
$n(n-2)(n+2)=0$

This equation yields three distinct integer solutions for $n$:
1) $n=0$
2) $n=2$
3) $n=-2$

Let us determine the three consecutive integers corresponding to each value of $n$ and verify them:
Case 1: For $n=0$, the integers are $-1$, $0$, and $1$.
Sum: $-1+0+1=0$
Product: $(-1)·0·1=0$ (Valid)

Case 2: For $n=2$, the integers are $1$, $2$, and $3$.
Sum: $1+2+3=6$
Product: $1·2·3=6$ (Valid)

Case 3: For $n=-2$, the integers are $-3$, $-2$, and $-1$.
Sum: $(-3)+(-2)+(-1)=-6$
Product: $(-3)·(-2)·(-1)=-6$ (Valid)

Thus, there are exactly three such possibilities.